Question

how do you find the slope of the tangent line?

Answer 1

Jim, you're good at explaining things...

Answer 2

Start with the slope of the secant line and draw the two points closer and closer together to get your approximated tangent line. This is all assuming you are not familiar with derivatives.

Answer 3

Is there a specific problem you're working on?

Answer 4

yes, I'm new to calc and don't know how to find the secant line and am not familiar with derivatives

Answer 5

its also assuming you know what a tangent line is too :)

Answer 6

\[\lim_{h->0}\frac{f(x+h)-f(x)}{h}\]

Answer 7

are you familiar with limits?

Answer 8

yes

Answer 9

gonna have to be if they want the tangent lines slope

Answer 10

so what is your function and where do you want to find the tangent line?

Answer 11

f(x)=sqrt(x) and tangent line at point 4,2

Answer 12

i still dont understand, i dont understand how you can insert 4,2 into one equation

Answer 13

Pleaseeeeee)

Answer 14

\[\large lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\]


\[\large lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\]


\[\large lim_{h \to 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\]


\[\large lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]


\[\large lim_{h \to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\]


\[\large lim_{h \to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\]


\[\large \frac{1}{\sqrt{x}+\sqrt{x}}\]


\[\large \frac{1}{2\sqrt{x}}\]


So \[\large lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{1}{2\sqrt{x}}\]

Answer 15

where does 4.2 come from? I'm trying to find the slope of the tangent line through the point 4, 2

Answer 16

So essentially, this shows that the derivative of \[\large f(x)=\sqrt{x}\] is \[\large f^{\prime}(x)=\frac{1}{2\sqrt{x}}\]


To find the slope of the line tangent to f(x) at x=4, simply plug in x=4 into the derivative function to get \[\large f^{\prime}(4)=\frac{1}{2\sqrt{4}}=\frac{1}{2*2}=\frac{1}{4}\]

Answer 17

So the slope of the tangent line is 1/4


Use this along with the given point to find the equation of the tangent line.

Answer 18

thanks, we haven't learned derivatives yet, so I'm still confused - but thanks

Answer 19

basically, that limit expression I showed above turns into the derivative function, which helps you find the slope of the tangent line

Answer 20

maybe I should just drop now before I dig too deep of a hole... thanks

Answer 21

don't worry, just get familiar with limits and the rest should be easier (since all of calculus is based on limits)

Answer 22

it's really the algebra that gets a lot of people